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AleksAgata [21]
4 years ago
8

One ordered pair $(a,b)$ satisfies the two equations $ab^4 = 12$ and $a^5 b^5 = 7776$. what is the value of $a$ in this ordered

pair?
Mathematics
1 answer:
LenaWriter [7]4 years ago
8 0
This is the concept of algebra, given that ab^4=12 and a^5b^5=7776, the value if a will be found as follows:
ab^4=12
a=12/b^4
also;
a^5=7776/b^5
thus;
a=(7776/b^5)^(1/5)
a=6/b
thus the value of a will be:
6/b=12/b^4
dividing both sides by b we get:
6=12/b^3
multiplying both sides by b^3 we get
6b^3=12
b^3=2
hence;
b=2^(1/3)



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8 0
3 years ago
4x – 2y = 2<br> x = y + 4​
Ray Of Light [21]

Answer:

y=-7; x=3

Step-by-step explanation:

You can substitute equation 2 into equation 1:

4(y+4)-2y=2

Then solve

4y+16-2y=2

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3 years ago
A rectangular storage container with an open top is to have a volume of 10 m3 . then length of its base is twice the width. mate
katrin [286]

Answer:

The cost of materials for the cheapest such container is $163.54.

Step-by-step explanation:

A rectangular storage container with an open top is to have a volume of 10 m³.

The volume of the rectangle is

\text{Volume} =\text{Length} \times \text{Width} \times \text{Height}

Length of its base is twice the width.

Let Width be 'w'.

Length is l=2w.

Height be 'h'.

10 =2w\times w\times h

10=2w^2h

The height in terms of width is represented as,

h=\frac{10}{2w^2}

h=\frac{5}{w^2}

According to question,

The cost is 10 times the area of the base and 6 times the total area of the sides.

i.e. Cost is given by,

C=10(L\times W)+6(2\times L\times H+2\times W\times H)

C=10(2w\times w)+6(2\times 2w\times \frac{5}{w^2}+2\times w\times \frac{5}{w^2})

C=20w^2+\frac{120}{w}+\frac{60}{w}

C(w)=20w^2+\frac{180}{w}

To get the minimum value,

Differentiate the cost w.r.t 'w',

C'(w)=20\frac{d(w^2)}{dw}+180\frac{d(w^{-1})}{dw}

C'(w)=20\times 2w-180 w^{-2}

C'(w)=40w-\frac{180}{w^2}

To find critical points put derivate =0,

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C''(w)=40\frac{d(w)}{dw}-180\frac{d(w^{-2})}{dw}

C''(w)=40+360(w^{-3})

C''(w)>0

As C''(w)>0 it is the minimum cost.

The cost is minimum at w=1.65.

Substitute the values in the cost function,

C(1.65)=20(1.65)^2+\frac{180}{1.65}

C(1.65)=54.45+109.09

C(1.65)=163.54

Therefore, the cost of materials for the cheapest such container is $163.54.

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